Suppose we have $x + y = 10$ what is the number of solutions for non-negative integers?
$$ 0 + 10 = 10 \\ 1 + 9 = 10 \\ 2 + 8 = 10 \\ 3 + 7 = 10 \\ 4 + 6 = 10 \\ 5 + 5 = 10 \\ 6 + 4 = 10 \\ 7 + 3 = 10 \\ 8 + 2 = 10 \\ 9 + 1 = 10 \\ 10 + 0 = 10 $$
we have 11 solutions.
Now Suppose we have $x + y + z = 10$ what is the number of solutions for non-negative integers?
$$ 0 + (x+y = 10) = 10 \ (11 \text{ ways}) \\ 1 + (x+y = 9) = 10 \ (10 \text{ ways})\\ 2 + (x+y = 8) = 10 \ (9 \text{ ways})\\ 3 + (x+y = 7) = 10 \ (8 \text{ ways})\\ 4 + (x+y = 6) = 10 \ (7 \text{ ways})\\ 5 + (x+y = 5) = 10 \ (6 \text{ ways})\\ 6 + (x+y = 4) = 10 \ (5 \text{ ways})\\ 7 + (x+y = 3) = 10 \ (4 \text{ ways})\\ 8 + (x+y = 2) = 10 \ (3 \text{ ways})\\ 9 + (x+y = 1) = 10 \ (2 \text{ ways})\\ 10 + (x+y = 0) = 10 \ (1 \text{ ways}) $$
we have $11 + 10 + \dots + 1 = \frac{(10 + 1)\times(10 + 2)}{2} = 66$ solutions.
Now Suppose we have $a + x + y + z = 10$ what is the number of solutions for non-negative integers?
$$ 0 + (x+y+z = 10) = 10 \ (66 \text{ ways}) \\ 1 + (x+y+z = 9) = 10 \ (55 \text{ ways})\\ 2 + (x+y+z = 8) = 10 \ (45 \text{ ways})\\ 3 + (x+y+z = 7) = 10 \ (35 \text{ ways})\\ 4 + (x+y+z = 6) = 10 \ (28 \text{ ways})\\ 5 + (x+y+z = 5) = 10 \ (21 \text{ ways})\\ 6 + (x+y+z = 4) = 10 \ (15 \text{ ways})\\ 7 + (x+y+z = 3) = 10 \ (10 \text{ ways})\\ 8 + (x+y+z = 2) = 10 \ (6 \text{ ways})\\ 9 + (x+y+z = 1) = 10 \ (3 \text{ ways})\\ 10 + (x+y+z = 0) = 10 \ (1 \text{ ways}) $$
we have $11 + 10 + \dots + 1 = \frac{(10 + 1)\times(10 + 2)\times(10 + 3)}{6} = 286$ solutions.
So I think the generalized solution for having sum $n$ given $m$ variables is $\frac{\frac{(n+m-1)!}{n!}}{(m-1)!}$
Am I on the right track?
